Redox Reactions and Electrochemistry

Problem Set

Chapter 5: 21-26, Chapter 21: 15-17, 32, 34, 43, 53, 72, 74

Oxidation/Reduction as Gain and Loss of Oxygen atoms

Oxidation – a reaction in which a substance gainsO atoms

(e.g. the oxidation of a hydrocarbon)

C

H

H

HR C

OH

H

HR C

O

HRC

O

OHRCO O OH2+

O2 O2 O2 O2

hydrocarbon alcohol aldehyde acid carbon dioxide

(i.e. this is equivalent to the combustion of a hydrocarbon)

Reduction - a reaction in which a substance losesO atoms

Oxidation/Reduction as Loss or Gain of Electrons

Cu(s)

Zn+2(aq)

Zn(s)

Cu+2(aq)

Cu+2(aq) + Zn(s) Cu(s) + Zn+2

(aq)

Presentation of Redox Reaction as 2 Half-Reactions

Cu+2(aq) + Zn(s) Cu(s) + Zn+2

(aq)

The reaction can be represented by two ½ reactions in which electrons are either gained or lost and the “oxidation state” of elements changes :

Cu+2(aq) + 2e- Cu(s) oxidation state of Cu: +2 0

Zn Zn+2(aq) + 2e- oxidation state of Zn: 0 +2

Reduction – a process in which electrons are gained (formally), the O.S. of an element decreases and electrons appear on the left side of the ½ reaction

Oxidation – a process in which electrons are lost (formally), the O.S. of an element increases and electrons appear on the right side of the ½ reaction

Oxidation states are governed by electronegativities - ability of atoms to compete for electrons with other atoms

More electronegative atoms are less willing to change their O.S.

Major Rules for Oxidation States1) O.S. of an atom in a free element is 0

2) Total of the O.S.’s of atoms in a molecule or ion is equal to the total charge on the molecule or ion.

3) Group 1A and Group 2A metals have O.S. of +1 and +2 respectively.

4) F always has an O.S. of –1. Cl also has O.S. of –1 unless it is bonded to oxygen or fluorine.

5) H almost always has an O.S. of +1.

6) O has O.S. of –2 (unless bonded to itself or F)

7) When bound to metals, group 7A, 6A and 5A elements have O.S.’s of -1, -2, -3 respectively.

Example of finding O.S.What is the formal O.S. of P in:

+1 -2

?

H3PO4 : 3 x (+1) + ? + 4 x (-2) = 0 (charge on molecule)

? = 8 - 3 = +5

H2PO4- : 2 x (+1) + ? + 4 x (-2) = -1 (charge on ion)

? = +5HPO4

2- : ? = +5 PO43- : ? = +5

If the oxidation state of elements do not change in a reaction, it is NOT a redox reaction

H3PO4(aq) + 3 OH-(aq) 3H2O + PO4

3-(aq) ACID/BASE Reaction

How to balance redox reactions?½ reaction method (not the same as method in textbook)

1) Identify species in which the oxidation state of an element is changing. Write the skelton ½ reactions including balancing of the redox atoms if necessary

2) Identify formal O.S. on both sides of equation for elements that have a change in O.S.

3) Add appropriate # of electrones to either left or right to balance oxidation states of redox atom(s).

4) Balance changes on left and right side of equation by adding H+ (if in acidic solution) or OH- if in basic solution.

5) Add appropriate number of H2O’s to left or right to balance atoms in the ½ reaction.

Balancing redox reactions, cont’dAt this point, both ½ reactions should be balanced. The next step is to combine the two ½ reactions to form an overall equation.

6) Multiply through each ½ reactions by appropriate coefficients to match e’s in each ½ reaction.

7) Add ½ reactions and cancel e’s and other common species on left and right.

8) Check Reaction. It should be balanced in terms of oxidation states, charge and atoms.

IF NOT, YOU HAVE MADE A MISTAKE!

Example of balancing a redox reactionDetermining sulphite in wastewater.

Sulphite is reacted with permanganate to produce sulphate and Mn(II) ion in acidic solution. Balance the redox reaction.

SO32- + MnO4

- SO42- + Mn2+ skeleton reaction

SO32- SO4

2- identify oxidation states

+4 +6

SO32- SO4

2- + 2e- balance O.S. with e-’s

SO32- SO4

2- + 2e- + 2H+ balance charges with H+

H2O + SO32- SO4

2- + 2e- + 2H+ balance atoms with H2O

Cont’d, Mn ½ reaction

MnO4- Mn2+ identify oxidation states

+7 +2

MnO4- + 5e- Mn2+ balance O.S. with e-’s

8 H+ + MnO4- + 5e- Mn2+ balance charges with H+

8 H+ + MnO4- + 5e- Mn2+ + 4H2O balance atoms with H2O

Balanced ½ Reactions

H2O + SO32- SO4

2- + 2e- + 2H+ x 5

8 H+ + MnO4- + 5e- Mn2+ + 4H2O x 2 to balance e’s

Cont’d, balancing full equationBalanced full reaction

16 H+ + 2MnO4- + 10e- 2Mn2+ + 8H2O x 2

5H2O + 5SO32- 5SO4

2- + 10e- + 10H+ x 56 3

5SO32- + 2MnO4

- + 6 H+ 5SO42- + 2Mn2+ + 3H2O

Check atom balance. OK

Try example 5.7 using this approach, use OH- to balance charge in basic solution. Much easier.

This method forces you to know O.S.

Example of balancing ½ reaction, that requires balancing redox atoms

Write the half reaction for Cr2O72- Cr+3

Cr2O72-

(aq) Cr+3(aq) skelton (in acidic solution),

Cr2O72-

(aq) 2 Cr+3(aq) determine the oxidation states

+6 +3Cr2O7

2-(aq) 2 Cr+3

(aq) balance redox atoms

Cr2O72-

(aq) + 2(3e-) 2 Cr+3(aq) balance O.S. with e-’s

Cr2O72-

(aq) + 6e- 2 Cr+3(aq) ” “

14 H+ + Cr2O72-

(aq) + 6e- 2 Cr+3(aq) balance charges with H+

14 H+ + Cr2O72-

(aq) + 6e- 2 Cr+3(aq) + 7H2O balance atoms with H2O

Disproportionation reactionsAn element in a substance is both oxidized and reduced

2 H2O2(aq) 2 H2O + O2(aq)

-1 -2 0

O OH

H Hydrogen peroxide: antiseptic agent, O2 acts as germicide

Ox/Red Agents

Oxidizing Agent – a chemical substance that oxidizes (removes electrons from) other substances in a chemical reaction. In the process of oxidizing something, the oxidant becomes reduced; it’s oxidation state decreases.

Reducing Agent – a chemical substance that reduces (loses electrons to) other substances. In the process of reducing, the oxidant becomes oxidized; it’s oxidation state increases.

Oxidizing and Reducing Agents

Loses

electrons

Oxidation half-reaction(reducing agent)

Reduction half-reaction(oxidizing agent)

Removes

electronsOxidation States of

Nitrogen

Oxidizing AgentsO2 – probably the most common and most important

oxidant known to us. Ubiquitous.

Organic Oxidation schemes (example- methane)

CH4

CHO

OH

CH2 OCH3 OH

O C O

-4 -2 0

+2

+4

O2 O2

O2

O2

Methane

(alkane)

Methanol

(alcohol)

formaldehyde

(aldehyde)

Formic acid

(carboxylic acid)

Carbon dioxide

(inorganic carbon)

Other oxidizing agentsOxides in their highest oxidation state are frequently strong oxidizing agents.

NO3 HNO3 NO2 NO N2O N2

+6 +5 +4 +2 +1 0

strong oxidizing agents weaker oxidizing agents

+7 +5 +3 +1

HClO4 ClO3- HClO2 HOCl

HNO3 , HClO4 – oxidizing acids

Non-oxidizing acids – HCl, HBr, HI, acids for which the only possible reduction ½ reaction is:

2H+(aq) + 2e- H2(g)

HNO3 is a much stronger oxidizing agent than H+.

Note: HNO3 has been used as an oxidant of a rocket fuel in military missiles

Cu, Ag, Au,Hg

Li, Na, K (1A metals)Mg, Ca (2A metals)Al, ZnFe, Sn, Pb

Metals that will not dissolve

Metals that dissolve in dilute H+ to produce H2

HNO3 dissolves Cu

Practice problem – Cu will dissolve in HNO3 producing Cu+2 in solution and a brown gas. Write a balanced equation for this process. (Hint: the brown gas is NO2)

ElectrochemistryCu(s) / Zn2+

(aq)

?

No reaction.

Cu(s) / Ag+(aq)

Cu2+/ Ag(s)

Spontaneous!

(∆G < 0)

Ag+(aq) + e- Ag(s) reduction

Cu(s) Cu2+(aq) + 2 e- oxidation

There is a flow of electrones from Cu to Ag+, but we cannon get use of this unless we have an electrical circuit

Not spontaneous!

(∆G > 0)

The reverse reaction is spont.

Electrochemical CellsFlow of electrons (current) can do work.

We can connect half reactions in separate containers through an electrical circuit. This will produce a current (electron flow) and voltage according to the spontaneity of the reactions.

Current depends on [ ], surface area of the electrodes, the resistance of the wires, etc.

V = 0.46 = const for this cell

Atomic view of a Voltaic (galvanic) Cell

Salt bridgeie – KNO3

Maintainsneutrality

Anode – oxidation Cathode - reduction

Cell Diagrams- anode (oxidation) is placed at left side of diagram- cathode (reduction) is placed at right side of diagram- boundary line, |, indicates a boundary between different phases (I.e. solution|solid)- a double boundary line || indicates a boundary (I.e.- salt bridge) between different half cell compartments

Zn(s) | Zn2+(aq) || Cu2+

(aq) | Cu(s)anodeoxidation

cathodereduction

half cell salt bridge half cell

Voltages and Current

+-anode cathode

Electromotive Force (EMF) - The voltage difference between two solutions provides a measure of the driving force of the electron transfer reaction.

Voltage is the difference of potentials at anode and cathode. How to determine potentials?

Standard Electrode PotentialsAbsolute measurements of potential (voltage) at a single point are meaningless, UNLESS, they are measured against some known reference.

In electricity, that reference is known as “ground”.

In electrochemistry, that reference is the standard hydrogen electrode (SHE).

A standard Electrode potential, Eo, measures the tendency for the reduction process to occur at an electrode (with respect to the SHE), when all species in solution have unit activity (~ 1.0 M) and gases are at 1 bar pressure.The higher E° the higher the tendency of reduction on the electrode (at the expence of oxidation at the electrode with lower E°).

Standard Hydrogen Electrode (SHE)

PH2 = 1.0 atm

aH3O+ = aH+ = 1.0

|| H+(aq) (~ 1M) | H2(g) 1 atm | Pt

2H+(aq) + 2e- H2(g)

Convention: The potential of the SHE is zero

EoH+(aq)/H2(g) = 0 V

Easy to reduce, hard to oxidize (good oxidizing agents)

↑ Oxidation is reverse of reduction:

E°oxidation =

- E°reduction

Hard to reduce, easy to oxidize (good reducing agents)↓

Standard Potential of Electrochemical CellThe potential of an electrochemical cell under standard conditions may be calculated as the sum of the reduction potential and oxidation potentials for two half cells:

Eocell = Eo

reduction + Eooxidation

Example 21-2: Reaction: Zn(s) + Cl2(g) ZnCl2(aq)

What is the standard potential of the cell, Eo.

Zn(s) Zn2+(aq) + 2e- Eo

Zn/Zn2+ = - EoZn2+/Zn = -(-0.763V) = +0.763 V

Cl2(g) + 2e- 2Cl-(aq) EoCl2/Cl- = + 1.358 V

Zn(s) + Cl2(g) ZnCl2(aq) Eocell = Eo

reduction + Eooxidation = + 2.121 V

E°oxidation

E°reduction

E°reduction

A simpler way of calculating E°cell

The potential of an electrochemical cell under standard conditions may be also calculated as Eo

cell = Eocathode – Eo

anode

where the Eo’s are standard reduction potentials taken from a table.

The definition of the cathode and anode depend on the direction of the reaction. Reduction occurs at the cathode (e-’s on L.S. of equation, O.S. decreasing) while oxidation occurs at the anode (e-’s on R.S of equation, O.S. increasing)

Also for a spontaneous reaction, Eocell > 0, as we will see

shortly.

Example 21-2 one more timeA new battery system currently under study for possible use in electric vehicles is the ZnCl2 battery.

Reaction: Zn(s) + Cl2(g) ZnCl2(aq)

What is the standard potential of the cell, Eo.

Zn(s) Zn2+(aq) + 2e- Eo

Zn2+/Zn = - 0.763 V Anode

Cl2(g) + 2e- 2Cl-(aq) EoCl2/Cl- = + 1.358 V Cathode

Zn(s) + Cl2(g) ZnCl2(aq)

Eocell = Eo

cathode – Eoanode = 1.358 – (- 0.763)V = 2.121 V

Reduction potentials

Spontaneous change in a CellPreviously, it was said Ecell > 0 for a spontaneous reaction. Where did this come from?

Electrical work:

Welectrical = Fl = Eql = {El = V} = qV F = electrical force, l = distance between two electrodes, q = charge moveed, V = voltage

If q in coloumbs and V in volts then W in joulesRelated parameters:

Power P = iV i = current (charge/time), V = voltage

If i - coloumbs/sec (Amp), V -volts, P- joules/sec = watts

In an electrochemical cell, q = n × F n = moles of electrons

F = charge of 1 mole of electrons = FaradayF = 96485 C/ mole electrons

V = Ecell

Spontaneous change, cont’dWelectrical = q V = nFEcell

This applies to a reversible process (implying that the reaction is carried out slowly enough that the system maintains equilibrium). Previously it was argued that the amount of work we can extract from a chemical process is equal –∆G (pg 796, Petrucci “Are You Wondering” box).

∆G = – Welectrical∆G = – nFEcell ∆Go = – nFEo

cell (standard molar → n is number of moles of electrons per mole of reaction)

If Eocell > 0, ∆Go < 0 and the reaction is spontaneous

If Eocell < 0, ∆Go > 0 and reaction is nonspontaneous

Behavior of Metals

Cu, Ag, Au,Hg

Li, Na, K (1A metals)Mg, Ca (2A metals)Al, ZnFe, Sn, Pb

Metals that will not dissolve

Metals that dissolve in dilute H+ to produce H2

Previously we said that experimental evidence shows the following:Metals of groups 1A and 2A are oxidized by H+.

Cu, Ag, Au, Hg are not oxidized by H+

Now we can better understand this:

M(s) Mn+(aq) + n e- oxidation

2 H+(aq) + 2 e- H2(g) reduction Eo = 0.00V

Eocell = Eo

cathode – Eoanode = Eo

H+(aq)/H2(g) – EoM+/M = 0 – Eo

M+/M

If EoM+/M < 0, Eo

cell > 0, the process is spontaneous

If EoM+/M > 0, a stronger oxidizing agent than H+ is required

(i.e. HNO3, HClO4) .

Oxidation of Gold in Aqua RegiaNitric acid, HNO3, is a powerful oxidizing agent but the chemical equilibrium for its reaction with gold, Au, only permits the formation of a very small amount of Au+3 ion:

Au(s) + 3NO3-(aq) + 6H+(aq) → Au3+(aq) + 3NO2(g) +3H2O(l)

so the amount of gold dissolved in pure nitric acid is very low.

We know that the equilibrium can be shifted to the right if the activity of Au3+ at the right is decreased by the formation of a complex ion, such as the chloraurate ion, AuCl4-:

Au3+(aq) + 4Cl-(aq) → AuCl4-

To form such an ion the solution has to contain Cl- ions. It is achieved though mixing HNO3 with HCl, this mixture is called aqua regia. Aqua regia can also dissolve platinum by a similar mechanisms.

←

←

Non-Standard Conditions:Nernst Equation

Previously, we have talked about standard electrode potentials, everything in standard state.

Very rarely are things in standard state.

In nonstandard state we use ∆G instead of ∆G°

∆G = ∆Go + RT ln Q R = gas constantT = temperature (K)Q = reaction quotient

-nFEcell = -nFEocell + RT ln Q

Ecell = Eocell – RT/nF ln Q = Eo

cell – RT/ (2.303 nF) log Q

For T = 25 °C RT/(2.303F) = const = 0.0592 V

C= −o ocell cell

0.0592E E logQ, for T = 25n Nernst Equation

Applications of the Nernst Equation

Draw the condensed cell diagram for the voltaic cell pictured at right.

Calculate the value of Ecell for T = 25°C

Pt | Fe2+(0.1M), Fe3+(0.2M) || Ag+(1.0M) | Ag(s)

The cell is in nonstandard conditions. We need to use the Nernst equation. Thus, we need to find Eo

cell , n, and Q.

1. To calculate E°cell it is enouigh to recognize half-reactions:From Table of Standard Reduction Potentials:Ag+ + e- Ag Eo

Ag+/Ag = 0.800 VFe3+ + e- Fe2+ Eo

Fe3+/Fe2+ = 0.771 VEo

cell: Eocell = Eo

cathode – Eoanode =

EoAg+/Ag – Eo

Fe3+/Fe2+= 0.800V – 0.771V = 0.029V2. To find both n and Q we need to write a balance overall reaction:

Cathode : Ag+ + e- Ag(s) (Reduction)Anode: Fe2+ Fe3+ + e- (Oxidation)

Overall: Ag+(aq) + Fe2+

(aq) + e- Fe3+(aq) + Ag(s) - e-

n = 1

unitless 21.0M0.10M

0.20M]][Ag[Fe

][FeQ 3

3

=×

== ++

+

Cont’d

The process is spontaneous under the stated non-standard conditions

0.011V log21

0.05920.029

logQn 2.303F

RTEE 0cellcell

=−=

=−=

Example A:Calculate Ecell at T = 25 °C for the following cell

Al|Al3+(0.36M) || Sn4+ (0.086 M), Sn2+ (0.54 M) |Pt

Sn4+ + 2e- Sn2+ EoSn4+/Sn2+ = 0.154 V

Al3+ + 3e- Al EoAl3+/Al = -1.676 V

Eocell = Eo

cathode – Eoanode = Eo

Sn4+/Sn2+ – EoAl3+/Al

= 0.154 – (-1.676) = 1.830 V

cont’d

Cathode: Sn4+ + 2e- Sn2+

Anode: Al Al3+ + 3e-

Overall: 3 Sn4+(aq) + 2 Al(s) + 6e- 3 Sn2+

(aq) + 2 Al3+(aq) - 6e-

+ +

+=2 3 3 2

4 3[Sn ] [Al ]Q

[Sn ]

x 3

x 2

n = 6

unitless 32.08(0.086M)

(0.36M)(0.54M)][Sn

][Al][SnQ 3

23

34

232

=== +

++ 3

0cell cell

RTE E logQ2.303F n

0.0592 1.830 log32.08 1.814 V6

= − =

= − =

The process is spontaneous under the stated non-standard conditions

Change in Ecell with Conditions

Zn(s) + Cu2+(aq) Zn2+

(aq) + Cu (s) T = 25 °C

E E 0.0592n

logQ = 1.100V 0.05922

log [Zn ][Cu ]

cell cello

2+

2+= − −

Slope = -59/2 mV per decade change

If we let cell reaction proceed, reaction shifts to right, [Zn2+] increases, [Cu2+] decreases and Ecelldecreases. When does it stop?

It stops at equilibrium, Ecell = 0.00V

0.0 E 0.0592n

logQcello

eq= − Keq !!

K 10eq

nE0.0592

cello

=

We can calculate Keqfrom Eo values! For above reaction, Keq = 1.5 x1037

Concentration CellsBoth ½ cells are the same chemical system, just different concentrations. The driving force (i.e. the EMF) is provided by the difference in concentrations.

Pt|H2 (g, 1.0 atm)| H+ (x M) || H+ (1 M) |H2(g,1.0 atm)|Pt(s)

Concentration CellCathode: 2 H+ (aq, 1 M) + 2e- H2 (g, 1 bar)

Anode: H2 (g, 1 bar) 2 H+ (xM) + 2e-

Overal: 2 H+ (aq, 1 M) 2 H+ (aq, x M) + 2e-

Eocell = Eo

cathode – Eoanode = 0.00 – 0.00 = 0.00V

( )

+

= − −

=− = − = −

x

2o anode

cell cell + 2cathode

2

2

[H ]0.0592 0.0592E E logQ= 0.00V logn 2 [H ]

0.0592 0.0592 log 2logx 0.0592logx2 1 2

Since pH = -log x, Ecell = 0.0592 pH

This concentration cell behaves as a pH meter! Other concentration cells can be used to measure unknown concentrations of other species (i.e. potentiometry).

Determination of Ksp (see example 21-10)From measured Ecell, determine Ksp of AgI.

Solution:

Set up the voltaic cell with saturated AgI, which produces A+ (aq, x M), at anode and Ag+ (aq, 0.1 M) at cathode, by dissolving AgNO3 for example.

Solve Nernst equation similar to how it was done in the previous example to get x.

x = [Ag+] = S, [I-] = S

Ag(s) | Ag+ (sat. AgI) || Ag+ (0.1M) |Ag(s)

Ksp = S2

Essentials of the last LectureNernst equation allows us to calculate the cell potential for non-standard conditions:

Ecell = Eocell – 2.303RT/nF log Q

Eocell is the standard potential of the cell

n is the number of moles of electrons per mole of balanced reaction

Q is the quation

Eocell is calculated from two half reactions

n and Q are found from the balanced overall redoxreaction

ElectrolysisThe use of an externally applied voltage to force an electrochemical reaction, even if it is naturally non-spontaneous. Spontaneous!

Zn(s) + Cu2+(aq) Zn2+

(aq) + Cu(s) Eocell = + 1.10 V

What about the reverse process? Non-spontaneous!

Zn2+(aq) + Cu(s) Zn(s) + Cu2+

(aq) Eocell = - 1.10 V

But if we apply a potential > 1.10 V across the cell, we overcome the natural negative voltage, thus providing the driving force to make the reaction proceed.

Current is in opposite direction of voltaic, or galvanic, cell.

Galvanic and Electrolytic CellsExternal energy External energy (voltage) source(voltage) source

electron flow reversed

Galvanic Cell Electrolytic CellRegardless of the cell type, anode and cathode always defined by the process: oxidation at the anode, reduction at the cathode.

Zn/Cu2+ electrolysis example continued...

The amount of current that flows in the electrolytic cell tells us how much Zn has been produced or how much Cu2+ has dissolved.

Faraday’s Law of Electrolysis:

The number of moles of product formed in an electrolysis cell by an electric current is chemically equivalent to the number of moles of electrons supplied.

or Q = nF = it

charge (coulombs) Total #

moles of electrons

current (amperes)

time (seconds)

Note: 1 A = 1C/s Faradays constant = 96485 C /mole of e-

n = it/F

Example 21-12Electrodeposition of Cu can be used to determine Cu2+ content of sample.

Cathode: Cu2+ + 2 e- Cu(s)

Anode: 2H2O O2(g) + 4H+(aq) + 4e-

What mass of Cu is deposited in 1 hr if current = 1.62A?

Solution: find moles of electrons, find moles of Cu, find mass of Cu

Mole of e- = = it/F = 1.62 A (C/s) × 3600 sec × 1/(96485 C/mole e-)

Mole of Cu = mole e- x 1 mole Cu / 2 mole e-

Mass Cu = moles Cu x 63.456 g Cu/mole Cu

Answer = 1.92 g of Cu deposited in 1 hour.

Cont’dExample B: How long will it take to produce 2.62 L of O2(g) at 26.2oC and 738 mmHg at a Pt anode with a constant current of 2.13A?

2H2O → 4H+ + O2(g) + 4e- oxidation of O on + electrode

4H+ + 4e- →2H2 (g) reduction of H on - electrode

Solution: find moles of O2, find moles of e-, find charge, find time.

Mole of O2:.

. .− −

×= =

×1 1

738 atm 2 62L760

0 08206Latmmol K 299 35KPVnRT

Mole of e- = moles of O2 x 4 mole e-/ mole O2

Charge = moles of e- × F (C/mole e-)

Time = Charge (C)/Current (C/s)

Answer = 18829 sec = 5.23 hr = 5 hr & 14min

BatteriesPrimary batteries (one use batteries) generate current through an irreversible reaction

Secondary batteries (rechargible batteries) generate current through a irreversible reaction.

Flow batteries serve as convertors of chemical energy to electrical while reactants (and products) flow through the batterie

Silver-Zinc Battery

1.8 V

Half reactions of discharge:

Reducation: Ag2O(s) + H2O(l) + 2e- → 2Ag(s) + 2OH-(aq)

Oxidation: Zn(s) + 2OH-(aq) → ZnO(s) + H2O(l) + 2e-

Overral: Zn(s) + Ag2O(s) → ZnO(s) + 2Ag(s)

The Nickel-Cadmium Rechargible Battery

1.4 V

Half reactions of discharge:

Reducation: 2NiO(OH)(s) + 2H2O(l) + 2e- → 2Ni(OH)2(s) + 2OH-(aq)

Oxidation: Cd(s) + 2OH-(aq) → Cd(OH)2(s) + 2e-

Overral: Cd(s) + 2NiO(OH)(s) + 2H2O → 2Ni(OH)2(s) + Cd(OH)2(s)

Hydrogen-Oxygen Flow CellHalf reactions of discharge:

Reducation: O2(g) + 2H2O(l) + 4e- → 4OH-(aq)

Oxidation: 2{H2(g) + 2OH-(aq) → 2H2O(l) + 2e-}Overral: 2H2(g) + O2(g) → 2H2O(s)

1.229 V